# Bound for a Sobolev function in an integral

For a compact bounded set $\Omega$, for the expression $$\int_\Omega \Delta u (\nabla u \cdot \nabla f)$$ where $u \in H^2$ and $f \in C^\infty$, is it possible to show that the expression is $\geq$ (greater than or equal to) some expression NOT involving $-\lVert \nabla u\rVert_{L^2}$? Notice the minus sign; with a plus sign it's fine. I can't see any way. So basically I want the expression $\geq \pm\lVert u \rVert_{L^2}$ or $\geq \lVert u \rVert_{H^2}$, with some multiplicative positive constants in the inequalities.

Any ideas appreciated.

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Is $f$ given or you can choose $f$ at your convenience? – timur Dec 26 '12 at 17:20
@timur $f$ is given. $f$ is bounded above and below by positive constants if that makes any difference. Just to clarify, I want to show the above for all $u \in H^2$. – Lemon Dec 26 '12 at 17:42
It looks like we need $f$ to be something special. If all we know is $1\le f\le 2$, then the devil can replace $f$ with $3-f$, and the integral changes sign. In such a situation the possible bounds for $\int$ are of the form $|\int|\le \|\dots\|$, which is what you do not want. – user53153 Dec 26 '12 at 19:15